Showing that $\sin(z+w)=\sin z\cos w+\cos z\sin w$, using the complex exponential definitions of sine and cosine

Question

Use the definitions $$\cos z=\frac{e^{iz}+e^{-iz}}2\qquad\sin z=\frac{e^{iz}-e^{-iz}}{2i}$$ to show that $$\sin(z+w)=\sin z\cos w+\cos z\sin w$$ I'm not quite sure on how to approach this question. My first attempt was substituting each of the terms on the right hand side with the definitions given i.e. letting $\sin z$ and $\cos z$ equal the definitions above, and then doing the same but substituting $w$ into the terms, but this didn't seem to lead me anywhere.

Answer 1

$$\sin z\cos w+\cos z\sin w=\dfrac{(e^{iz}-e^{-iz})(e^{iw}+e^{-iw})+(e^{iz}+e^{-iz})(e^{iw}-e^{-iw})}{4i}$$

$\dfrac{\text{The numerator}}2$ $$=e^{i(z+w)}-e^{-i(z+w)}+e^{i(z-w)}-e^{i(w-z)}-e^{i(z-w)}+e^{i(w-z)}$$